How many grams of solid calcium hydroxide, Ca(OH)2, are required to react with 350 mL of 0.40 M HCl? __HCl + __Ca(OH)2
__CaCl2 + __H2O

Let's determine how many moles there are based on the chemical reactions;

NOW THAT THE MOLARITY AND VOLUME OF HYDROCHLORIC ACID ARE GIVEN, THE MASS OF HYDROCHLORIC ACID IN THE SOLUTION CAN BE CALCULATED.

MOLES of HCl; volume of solution x molarity of the solution

Number of moles of HCl = 350 mL x 0.40 mol/L

= 0.14 mol/L x 0.350 L (0.40 mol/L)

number of moles times the molar mass of HCl is the mass of HCl that yields 0.14 mol of HCl.

36.5 g/mol x 0.14 mol is the mass of HCl.

HCl mass equals 5.11g.

To find the number of grams of Ca(OH)2 required:

1. Calculate the number of moles of HCl using the formula: moles = Molarity × Volume (in liters).
2. Use the balanced chemical equation to determine the mole ratio between HCl and Ca(OH)2.
3. Calculate the number of moles of Ca(OH)2 needed based on the mole ratio.
4. Convert the moles of Ca(OH)2 to grams using the molar mass of Ca(OH)2.

To find the amount of solid calcium hydroxide (Ca(OH)₂) required to react with 350 mL of 0.40 M hydrochloric acid (HCl), we first need to determine the balanced chemical equation for the reaction between HCl and Ca(OH)₂.

The balanced equation is: 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

From the balanced equation, we see that 2 moles of HCl react with 1 mole of Ca(OH)₂.

Next, we calculate the moles of HCl present in 350 mL of 0.40 M HCl: (0.40 \text{ moles/L} \times 0.350 \text{ L} = 0.14 \text{ moles}) of HCl

Since the ratio of HCl to Ca(OH)₂ is 2:1, the moles of Ca(OH)₂ required will be half of the moles of HCl: (0.14 \text{ moles} \div 2 = 0.07 \text{ moles}) of Ca(OH)₂

Now, we calculate the molar mass of Ca(OH)₂: Ca: 1 atom x 40.08 g/mol = 40.08 g/mol O: 2 atoms x 16.00 g/mol = 32.00 g/mol H: 2 atoms x 1.01 g/mol = 2.02 g/mol Total molar mass = 40.08 + 32.00 + 2.02 = 74.10 g/mol

Finally, we calculate the mass of Ca(OH)₂ required: (0.07 \text{ moles} \times 74.10 \text{ g/mol} = 5.19 \text{ grams})

Therefore, 5.19 grams of solid calcium hydroxide are required to react with 350 mL of 0.40 M HCl.